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Thus, we can see a fine balance here. 04988, well below the content size s = 100. Let us study in some detail the optimal merging limit. 8). Letting λ → 0, we have x0 (λ ) → s, which means for a smaller request rate we cannot count on the following requests, so “be a child whenever you can” is the best strategy. On the contrary, for large λ , we have x0 (λ ) → 0, as λ → ∞. For a larger request rate, you can always expect the following requests. ” If we use unicast only, instead of the combination of unicast and multicast, the average streaming rate is ρ .

Assume the requests to the content of size s is a Poisson process with the rate λ . Given the merging limit time x, the average download rate is obtained by b(x) = 2λ s + λ 2 x2 . 7) The function b(x) is indeed a convex function (see Fig. 4), so we have x0 which minimizes b(x) as x0 = (1 + 2λ s)1/2 − 1 . 9) where ρ = λ s corresponds to the scale of this streaming service. Proof. We know that S(t) is a renewal reward process from Sect. 2. 1] is an extension of the strong law of large numbers to the renewal process.